✔ 最佳答案
圖片參考:
http://imgcld.yimg.com/8/n/AD04686329/o/151006150056813871800250.jpg
(Note: F為BE直線與圓的交點,請自行補上EF線段)
1. ∆PAB:∆PAD:∆PAM=PA:PD:PM
2. PA+PD=2PE (E為圓心O至AD之垂足)
3. (重點)PBEC四點共圓
因∆PBC之外接圓即OP之直徑之圓
又∠OEP=90度, so, E為∆PBC之外接圓上
故∠CPB=∠CBF
即(CF弧+FD弧-AC弧)/2=CF弧/2 or AC弧=DF弧
4. PA*PD=PB^2 (切割線性質)
5. 1/∆PAB+ 1/∆PAD= 2/∆PAM
<=> 1/PA+ 1/PD= 2/PM
<=> 2PE/PB^2= 2/PM
<=> PB/PM = PE/PM
<=> ∆PBM~∆PEB
<=> ∠PBM=∠PEB
<=> ∠PBA+∠ABM=∠EDB+∠EBD
<=> ∠ABM=∠EBD
<=> AC弧=DF弧
(Note:第5步驟倒敘即得證)
2010-06-15 03:05:30 補充:
5. 1/∆PAB+ 1/∆PAD= 2/∆PAM
<=> 1/PA+ 1/PD= 2/PM
<=> 2PE/PB^2= 2/PM
<=> PB/PM = PE/PB
<=> ∆PBM~∆PEB
<=> ∠PBM=∠PEB
<=> ∠PBA+∠ABM=∠EDB+∠EBD
<=> ∠ABM=∠EBD
<=> AC弧=DF弧
2010-06-15 23:44:19 補充:
To:老王
請問:圓對稱的調和性質,是什麼幾何定理?
我可是證明得很辛苦,感覺繞得很遠!
2010-06-16 18:58:58 補充:
謝謝老王!